题目：
Given a binary search tree and a node in it, find the in-order successor of that node in the BST.
Note: If the given node has no in-order successor in the tree, return null.
链接： http://leetcode.com/problems/inorder-successor-in-bst/
题解：
一开始的想法就是用inorder traversal，设置一个boolean变量，当找到root.val = p.val的时候返回下一个节点，遍历完毕以后返回null。
Time Complexity - O(n)， Space Complexity - O(n)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { if(root == null || p == null) { return null; } boolean foundNodeP = false; Stack<TreeNode> stack = new Stack<>(); while(root != null || !stack.isEmpty()) { if(root != null) { stack.push(root); root = root.left; } else { root = stack.pop(); if(foundNodeP) { return root; } if(root.val == p.val) { foundNodeP = true; } root = root.right; } } return null; } }
看了Discuss之后发现有很多简洁的写法，而且不用遍历全部元素。利用BST的性质，比较root.val和p.val，然后在左子树或者右子树中查找。
Time Complexity - O(h)， Space Complexity - O(1)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { if(root == null || p == null) { return null; } TreeNode successor = null; while(root != null) { if(p.val < root.val) { successor = root; root = root.left; } else { root = root.right; } } return successor; } }
Reference:
https://leetcode.com/discuss/69200/for-those-who-is-not-so-clear-about-inorder-successors
https://leetcode.com/discuss/61105/java-python-solution-o-h-time-and-o-1-space-iterative
https://leetcode.com/discuss/59728/10-and-4-lines-o-h-java-c
https://leetcode.com/discuss/59787/share-my-java-recursive-solution
原文：http://www.cnblogs.com/yrbbest/p/5037889.html